Base | Representation |
---|---|
bin | 1001000011010111001… |
… | …01001011101111110111 |
3 | 1002201212002102221021212 |
4 | 10201223211023233313 |
5 | 20044003343223441 |
6 | 354520241232035 |
7 | 31320632044016 |
oct | 4415345135767 |
9 | 1081762387255 |
10 | 311042554871 |
11 | 10aa04460204 |
12 | 5034763261b |
13 | 2343c805949 |
14 | 110a992ca7d |
15 | 8156d56eeb |
hex | 486b94bbf7 |
311042554871 has 2 divisors, whose sum is σ = 311042554872. Its totient is φ = 311042554870.
The previous prime is 311042554829. The next prime is 311042554873. The reversal of 311042554871 is 178455240113.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 311042554871 - 218 = 311042292727 is a prime.
Together with 311042554873, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (311042554873) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155521277435 + 155521277436.
It is an arithmetic number, because the mean of its divisors is an integer number (155521277436).
Almost surely, 2311042554871 is an apocalyptic number.
311042554871 is a deficient number, since it is larger than the sum of its proper divisors (1).
311042554871 is an equidigital number, since it uses as much as digits as its factorization.
311042554871 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 134400, while the sum is 41.
Adding to 311042554871 its reverse (178455240113), we get a palindrome (489497794984).
The spelling of 311042554871 in words is "three hundred eleven billion, forty-two million, five hundred fifty-four thousand, eight hundred seventy-one".
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